The solubility product expression for $\mathrm{Ca}_3\left(\mathrm{PO}_4\right)_2$ is represented as

The solubility product expression for $\mathrm{Ca}_3\left(\mathrm{PO}_4\right)_2$ is represented as
  1. $\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ca}^{2+}\right]^2\left[\mathrm{PO}_4{ }^{3-}\right]^2$
  2. $\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ca}^{2+}\right]^3\left[\mathrm{PO}_4{ }^{3-}\right]^2$
  3. $\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ca}^{2+}\right]\left[\mathrm{PO}_4{ }^{2-}\right]^3$
  4. $\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ca}^{2+}\right]\left[\mathrm{PO}_4{ }^{2-}\right]$

Solution

$\begin{aligned} & \mathrm{Ca}_3\left(\mathrm{PO}_4\right)_{2(\mathrm{~s})} \rightleftharpoons 3 \mathrm{Ca}_{(\mathrm{aq})}^{2+}+2 \mathrm{PO}_{4(\mathrm{aq})}^{3-} \\ & \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ca}^{2+}\right]^3\left[\mathrm{PO}_4^{3-}\right]^2\end{aligned}$

Asked in: MHT CET 2021 (24 Sep Shift 2)

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